Answer:
Cesium Carbonate
Explanation:
2CsOH + H2CO3 → Cs2CO3 + 2H2O
Answer:
The answer is "Option b and Option c".
Explanation:
This buffer is a buffer of ammonia and ammonium ion. Thus it requires the solution
.
In point 1:
The solution containing
at 1M concentration would be given by mixing the two solutions. Thus, this buffer is a legitimate route.
In point 2:
It gives the ions you want but they are not the same.
In point 3:
and 
volume would not produce the same
concentrations. Therefore, this buffer isn't a valid route.
In point 4:
Some
volume and half
. This offers the same rate as half.
The molarity is the number of moles in 1 L of the solution.
The mass of NH₃ given - 2.35 g
Molar mass of NH₃ - 17 g/mol
The number of NH₃ moles in 2.35 g - 2.35 g / 17 g/mol = 0.138 mol
The number of moles in 0.05 L solution - 0.138 mol
Therefore number of moles in 1 L - 0.138 mol / 0.05 L x 1L = 2.76 mol
Therefore molarity of NH₃ - 2.76 M
Initial moles of C₆H₅COOH = 500/1000 × 0.10 = 0.05mol
Initial moles of C₆H₅COONa = 500/1000 × 0.10 = 0.05 mol
initial pH = Pka + log([C₆H₅COONa/ moles of C₆H₅COOH)
4.19 = pKa + log(0.05/0.05)
→pKa = 4.19
C₆H₅COOH + NaOH → C₆H₅COONa ₊ H₂o
moles of NaOH added = 0.010 mol
moles of C₆H₅COOH = 0.05 - 0.025 = 0.025 mol
Final pH = pKa + log([C₆H₅COONa)/[ C₆H₅COOH])
=pKa + log(moles of C₆H₅COONa/moles of C₆H₅COOH)
= 4.19 + log(0.025/0.075)
4.29
We first need to find the number of moles of gas in the container
PV = nRT
where;
P - pressure - 2.87 atm x 101 325 Pa/atm = 290 802.75 Pa
V - volume - 5.29 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 230 K
substituting these values in the equation
290 802.75 Pa x 5.29 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 230 K
n = 0.804 mol
the molar mass = mass present / number of moles
molar mass of gas = 56.75 g / 0.804 mol
therefore molar mass is 70.6 g/mol